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cos3x = 4cos^3 x - 3cosx?
L.H.S. = cos3x = cos(2x+x) = cos2x.cosx - sin2x.sinx
L.H.S. = (2cos^2x -1)cosx - 2sinx.cosx.sinx
L.H.S. = 2cos^3x -cosx - 2(1-cos^2x)cosx
L.H.S. = 2cos^3x -cosx -2cosx + 2cos^3x
L.H.S. = 4cos^3x - 3cosx = R.H.S. >=========< Q . E . D.
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cos(3x) = cos(2x + x) = cos(2x)cos(x) – sin(2x)sin(x) =
(2cos²x – 1)(cosx) – 2sinx cosx sinx = cosx[(2cos²x – 1) – 2(1 – cos²x)] =
2cos³x – cosx – 2cosx + 2cos³x = 4cos³x – 3cosx QED
By the cosine addition formula, we can write:
LHS = cos(3x)
= cos(2x + x)
= cos(2x)cos(x) - sin(2x)sin(x), since cos(A +
= cos(A)cos(B) - sin(A)sin(B)
= cos(x)[2cos^2(x) - 1] - [2sin(x)cos(x)]sin(x), by the double-angle formulas
= 2cos^3(x) - cos(x) - 2sin^2(x)cos(x), by expanding
= 2cos^3(x) - cos(x) - 2cos(x)[1 - cos^2(x)], since sin^2(x) + cos^2(x) = 1
= 2cos^3(x) - cos(x) - [2cos(x) - 2cos^3(x)]
= 2cos^3(x) - cos(x) - 2cos(x) + 2cos^3(x)
= 4cos^3(x) - 3cos(x)
= RHS.
I hope this helps!
cos3x = cos (2x+x)
= cos (2x)*cos (x) - sin (2x)*sin (x)
= (2cos^2 (x) - 1)cos (x) - 2sin^2 (x)*cos (x)
= 2cos^3 (x) - cos (x) - 2(1 - cos^2 (x))cos (x)
= 2cos^3 (x) - cos (x) - 2cos (x) + 2cos^3 (x)
= 4cos^3 (x) - 3cos (x)
= R.H.S